[R] Calculate daily means from 5-minute interval data

Richard O'Keefe r@oknz @end|ng |rom gm@||@com
Tue Aug 31 07:11:18 CEST 2021

By the time you get the data from the USGS, you are already far past the point
where what the instruments can write is important.
(Obviously an instrument can be sufficiently broken that it cannot
write anything.)
The data for Rogue River that I just downloaded include this comment:

# Data for the following 1 site(s) are contained in this file
# -----------------------------------------------------------------------------------
# Data provided for site 04118500
#            TS   parameter     Description
#         71932       00060     Discharge, cubic feet per second
# Data-value qualification codes included in this output:
#     A  Approved for publication -- Processing and review completed.
#     P  Provisional data subject to revision.
#     e  Value has been estimated.
agency_cd site_no datetime tz_cd 71932_00060 71932_00060_cd
5s 15s 20d 6s 14n 10s

(I do not know what the last line signifies.)
It is, I think, sufficiently clear that the instrument does not know what
the qualification code is!

After using read.delim to read the file

I note that the timestamps are in a single column, formatted like
"2020-08-30 00:15", matching the pattern "%Y-%m-%d %H:%M".

After reading the data into R and using
r$datetime <- as.POSIXct(r$datetime, format="%Y-%m-%d %H:%M",
I get
  agency           site            datetime                     tz
 USGS:33550   Min.   :4118500   Min.   :2020-08-30 00:00:00   EST:33550
              1st Qu.:4118500   1st Qu.:2020-11-25 13:33:45
              Median :4118500   Median :2021-03-08 03:52:30
              Mean   :4118500   Mean   :2021-03-01 07:05:54
              3rd Qu.:4118500   3rd Qu.:2021-06-03 12:41:15
              Max.   :4118500   Max.   :2021-08-30 22:00:00
      flow        qual
 Min.   : 96.5   A  :18052
 1st Qu.:156.0   A:e:  757
 Median :193.0   P  :14741
 Mean   :212.5
 3rd Qu.:237.0
 Max.   :767.0

So for this data set, spanning one year, all the times are in the same time
zone, observations are 15 minutes apart, not 5, and there are no missing
data.  This was obviously the wrong data set.
Oh well, picking an epoch such as
> epoch <- min(r$datetime)
and then calculating
as.numeric(difftime(timestamp, epoch, units="min")))
will give you a minute count from which determining day number
and bucket within day is trivial arithmetic.

I have attached a plot of the Rogue River flows which should make it
very clear what I mean by saying that means and standard deviations
are not a good way to characterise this kind of data.

The flow is dominated by a series of "bursts" with a fast onset to a peak
and a slow decay, coming in a range of sizes from quite small to rather
large, separated by gaps of 4 to 45 days.

I'd be looking at
 - how do I *detect* these bursts? (detecting a peak isn't too hard,
   but the peak is not the onset)
 - how do I *characterise* these bursts?
   (and is the onset rate related to the peak size?)
 - what's left after taking the bursts out?
 - can I relate these bursts to something going on upstream?

My usual recommendation is to start with things available in R out of the
box in order to reduce learning time.

On Tue, 31 Aug 2021 at 11:34, Rich Shepard <rshepard using appl-ecosys.com> wrote:
> On Tue, 31 Aug 2021, Richard O'Keefe wrote:
> > I made up fake data in order to avoid showing untested code. It's not part
> > of the process I was recommending. I expect data recorded every N minutes
> > to use NA when something is missing, not to simply not be recorded. Well
> > and good, all that means is that reshaping the data is not a trivial call
> > to matrix(). It does not mean that any additional package is needed or
> > appropriate and it does not affect the rest of the process.
> Richard,
> The instruments in the gauge pipe don't know to write NA when they're not
> measuring. :-) The outage period varies greatly by location, constituent
> measured, and other unknown factors.
> > You will want the POSIXct class, see ?DateTimeClasses. Do you know whether
> > the time stamps are in universal time or in local time?
> The data values are not timestamps. There's one column for date a second
> colume for time and a third column for time zone (P in the case of the west
> coast.
> > Above all, it doesn't affect the point that you probably should not
> > be doing any of this.
> ? (Doesn't require an explanation.)
> Rich
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.

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